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Unformatted text preview: wait for an
exponential amount of service with rate µ2 . - µ1 - µ2 - Our main problem is to ﬁnd conditions that guarantee that the queue stabilizes, i.e., has a stationary distribution. This is simple in the tandem queue.
The ﬁrst queue is not a↵ected by the second, so if &lt; µ1 , then (4.23) tells us
that the equilibrium probability of the number of customers in the ﬁrst queue,
1
Xt , is given by the shifted geometric distribution
✓ ◆m ✓
◆
1
P (Xt = m) =
1
µ1
µ1
In the previous section we learned that the output process of an M/M/1
queue in equilibrium is a rate Poisson process. This means that if the ﬁrst
2
queue is in equilibrium, then the number of customers in the queue, Xt , is itself
an M/M/1 queue with arrivals at rate (the output rate for 1) and service rate
µ2 . Using the results in (4.23) again, the number of individuals in the second
queue has stationary distribution
✓ ◆n ✓
◆
2
P (Xt = n) =
1
µ2
µ2
To specify the stationary distribution of the system, we need to know the
1
2
joint distribution of Xt and Xt . The answer is somewhat remarkable: in 144 CHAPTER 4. CONTINUOUS TIME MARKOV CHAINS equilibrium the two queue lengths are independent.
✓ ◆m ✓
◆ ✓ ◆n ✓
1
2
P (Xt = m, Xt = n) =
1
·
1
µ1
µ1
µ2 µ2 ◆ (4.27) Why is this true? Theorem 4.11 implies that the queue length and the depart...
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